Question

实数 $x$ 和 $y$ 独立且均匀地从区间 $[0, 1]$ 中随机选取。下面哪个数最接近 $x$ 、 $y$ 和 $1$ 能作为一个钝角三角形三边长的概率？

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $[0, 1].$ Which of the following numbers is closest to the probability that $x,$ $y,$ and $1$ are the side lengths of an obtuse triangle?

$0.21$

$0.25$

$0.29$

$0.50$

$0.79$

答案：C

知识点：几何概率余弦定理

难度评级：2100

解答：

三边 $x, y, 1$ 构成三角形，当且仅当 $x + y > 1$ 又因为 $1$ 是最长边，该三角形为钝角，当且仅当 $x^2 + y^2 < 1$ 因此在单位正方形中，所求区域位于四分之一圆 $x^2 + y^2 = 1$ 内、直线 $x + y = 1$ 上方，面积为 $\tfrac{\pi}{4} - \tfrac12 \approx 0.285$ 最接近的选项是 $0.29$ 正确答案是 C。

The three lengths $x, y, 1$ make a triangle iff $x + y > 1.$ Since $1$ is the longest side, that triangle is obtuse iff $x^2 + y^2 < 1.$ So in the unit square we want the region inside the quarter circle $x^2 + y^2 = 1$ but above the line $x + y = 1.$ That's the quarter disk with the right triangle under the chord removed: $\tfrac{\pi}{4} - \tfrac12 \approx 0.285.$ The closest choice is $0.29.$ Therefore, the answer is C.

2018 AMC 10B 第 22 题

答案：C

难度评级：2100

解答：

其他年份的第 22 题